Quoted post said:Originally Posted by Jim Moore [IMG]/img/forum/go_quote.gif[/IMG]
Here's some stuff I jsut found on the net. Man, you ain't kidding about wind affecting speed. From my calculations it looks like you drop from 20 mph to about 17 mph riding into a 5 mph headwind. that's more of a drop than I expected but it explains why I suck so bad in the wind. Can someone check my math?
Power requiredThere is a well known equation that gives the power required to push a bike/rider through the air and to overcome the friction of the drive train:
P=gmVg (.0053) + VgVaVa(.185)
Where P is in watts, g is Earth's gravity, Vg is ground speed (m/s), m is bike/rider mass in kg, s is the grade (m/m), and Va is the rider's speed through the air (m/s). K1 is a lumped constant for all frictional losses (tires, bearings, chain), and is generally reported with a value of 0.0053. K2 is a lumped constant for aerodynamic drag and is generally reported with a value of 0.185 kg/m.
I used 82kg and (1 mph = 0.45 m/s).
No.
First, the units are all wrong, assuming the constants (K1 and K2) are unitless. Units of power are kg∙(m^2)/(s^3), i.e. watts. Your first term yields units of kg∙(m^3)/(s^5), while your second term gives units of (m^3)/(s^3).
Second, in a simple model you would use the resultant velocity, V=Vb+Vw[cos(θ)], where Vb is the velocity of the bike, Vw is the velocity of the wind, and θ is the angle between the direction of bike travel and the wind. If the wind is blowing in a direction opposite the bike, Vw is negative. Likewise, if the wind is blowing in the same direction of bike travel, Vw is positive.
Third, the forces to overcome will be:
--bearing drag (can apply to chain too) and rolling resistance
--aero drag
--gravity (only if on a slope)
Power relates to force by P=F∙V.
The power needed to overcome gravity will then be P=mgVb[sin(θ)], where m is mass, g is the acceleration of gravity, and θ is the angle of the road to the horizontal.
The power needed to overcome drive train losses, bearing losses, and rolling resistance will be P=mgVb(Crr+Ct+Cb)cos(θ), where Crr is the coefficient of rolling resistance, Ct is the coefficient of drive train resistance, and Cb would be coefficient of bearing resistance.
The power needed to overcome aero drag will be P=ÏACdV^3, where Ï is the air density, A is the area presented to the wind, Cd is the coefficient of drag, and V is as defined in the second paragraph.
Again, that's a simple model. Cd will actually change depending on the direction of the wind and the position of the rider on the bike.
Perhaps the best thing to do, if you really want to do this the simple way, is use Analytic Cycling's "Wind on Rider" calculator to see how different performance metrics vary with wind speed and direction. Here's the link: http://www.analyticcycling.com/DiffEqWindCourse_Page.html .